A162514 -id:A162514 - cp's OEIS frontend

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A162517 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x + d)^n - (x - d)^n)/(2*d), where d = sqrt(x+4).

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 4, 4, 16, 0, 5, 10, 41, 8, 16, 6, 20, 86, 48, 96, 0, 7, 35, 161, 169, 348, 48, 64, 8, 56, 280, 456, 992, 384, 512, 0, 9, 84, 462, 1044, 2449, 1744, 2400, 256, 256, 10, 120, 732, 2136, 5482, 5920, 8640, 2560, 2560, 0, 11, 165, 1122, 4026, 11407, 16721, 26420, 14240, 14720, 1280, 1024

Offset: 1

Clark Kimberling, Jul 05 2009

			First six rows:
  0
  1
  2...0
  3...1...4
  4...4...16...0
  5...10..41...8...16

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Cf. A162514, A162515, A162516.

Cf. A002605, A063727.

m:=12;
Q:= func< n,x | ((x+Sqrt(x+4))^n - (x-Sqrt(x+4))^n)/(2*Sqrt(x+4)) >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( Q(n, x) ), n-k) >;
[0] cat [T(n,k): k in [1..n], n in [1..m]]; // G. C. Greubel, Jul 09 2023

Q[n_, x_]:= Q[n, x]= ((x+Sqrt[x+4])^n -(x-Sqrt[x+4])^n)/(2*Sqrt[x+4]);
T[n_, k_]:= Coefficient[Series[P[n,x], {x,0,n-k+1}], x, n-k];
Join[{0}, Table[T[n,k], {n,12}, {k,n}]//Flatten] (* G. C. Greubel, Jul 09 2023 *)

def Q(n,x): return ((x+sqrt(x+4))^n - (x-sqrt(x+4))^n)/(2*sqrt(x+4))
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( Q(n,x) ).list()[n-k]
[0]+flatten([[T(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Jul 09 2023

Q(n,x) = (P(n+1, x) - x*P(n,x))/(x+4), where P(n, x) is the n-th polynomial of A162516.
Q(n, x) also has the recurrence Q(n, x) = 2*x*Q(n-1, x) - (x^2 - x - 4)*Q(n-2, x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)](((x+sqrt(x+4))^n -(x-sqrt(x+4))^n)/(2*sqrt(x+4))).
Sum_{k=1..n-1} T(n, k) = A063727(n-2), n >= 2.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A002605(n-1). (End)

A162515 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = (U^n - L^n)/d, where U = (x + d)/2, L = (x - d)/2, d = sqrt(x^2 + 4).

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 3, 0, 1, 1, 0, 4, 0, 3, 0, 1, 0, 5, 0, 6, 0, 1, 1, 0, 6, 0, 10, 0, 4, 0, 1, 0, 7, 0, 15, 0, 10, 0, 1, 1, 0, 8, 0, 21, 0, 20, 0, 5, 0, 1, 0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 1, 0, 10, 0, 36, 0, 56, 0, 35, 0, 6, 0, 1, 0, 11, 0, 45, 0, 84, 0, 70, 0, 21, 0, 1

Offset: 0

Clark Kimberling, Jul 05 2009

Row sums 0,1,1,2,3,5,... are the Fibonacci numbers, A000045.
Note that the coefficients are given in decreasing order. - M. F. Hasler, Dec 07 2011
Essentially a mirror image of A168561. - Philippe Deléham, Dec 08 2013

			Polynomial expansion:
  0;
  1;
  x;
  x^2 + 1;
  x^3 + 2*x;
  x^4 + 3*x^2 + 1;
First rows:
  0;
  1;
  1, 0;
  1, 0, 1;
  1, 0, 2, 0;
  1, 0, 3, 0, 1;
  1, 0, 4, 0, 3, 0;
Row 6 matches P(6,x)=x^5 + 4*x^3 + 3*x.

G. C. Greubel, Rows n = 0..101 of triangle
T. Copeland, Addendum to Elliptic Lie Triad

Cf. A000045, A049310, A053119, A162514, A162516, A162517.

T:= function(n,k)
    if (k mod 2)=0 then return Binomial(n- k/2, k/2);
    else return 0;
    fi; end;
Concatenation([0], Flat(List([0..15], n-> List([0..n], k-> T(n,k) ))) ); # G. C. Greubel, Jan 01 2020

function T(n,k)
  if (k mod 2) eq 0 then return Round( Gamma(n-k/2+1)/(Gamma(k/2+1)*Gamma(n-k+1)));
  else return 0;
  end if; return T; end function;
[0] cat [T(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 01 2020

0, seq(seq(`if`(`mod`(k,2)=0, binomial(n-k/2, k/2), 0), k = 0..n), n = 0..15); # G. C. Greubel, Jan 01 2020

Join[{0}, Table[If[EvenQ[k], Binomial[n-k/2, k/2], 0], {n,0,15}, {k,0,n} ]//Flatten] (* G. C. Greubel, Jan 01 2020 *)

row(n,d=sqrt(1+x^2/4+O(x^n))) = Vec(if(n,Pol(((x/2+d)^n-(x/2-d)^n)/d)>>1)) \\ M. F. Hasler, Dec 07 2011, edited Jul 05 2021

@CachedFunction
def T(n, k):
    if (k%2==0): return binomial(n-k/2, k/2)
    else: return 0
[0]+flatten([[T(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jan 01 2020

P(n,x) = x*P(n-1, x) + P(n-2, x), where P(0,x)=0 and P(1,x)=1.

T(n,k) = T(n-1, k) + T(n-2, k-2) for n>=2. - Philippe Deléham, Dec 08 2013

A162516 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x+d)^n + (x-d)^n)/2, where d=sqrt(x+4).

Original entry on oeis.org

1, 1, 0, 1, 1, 4, 1, 3, 12, 0, 1, 6, 25, 8, 16, 1, 10, 45, 40, 80, 0, 1, 15, 75, 121, 252, 48, 64, 1, 21, 119, 287, 644, 336, 448, 0, 1, 28, 182, 588, 1457, 1360, 1888, 256, 256, 1, 36, 270, 1092, 3033, 4176, 6240, 2304, 2304, 0, 1, 45, 390, 1890, 5925, 10801, 17780, 11680, 12160, 1280, 1024

Offset: 0

Clark Kimberling, Jul 05 2009

			First six rows:
  1;
  1,  0;
  1,  1,  4;
  1,  3, 12,  0;
  1,  6, 25,  8, 16;
  1, 10, 48, 40, 80, 0;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A162514, A162515, A162517.
Cf. A000217, A026150, A084057, A125818, A199572.
For fixed k, the sequences P(n,k), for n=1,2,3,4,5, are A084057, A084059, A146963, A081342, A081343, respectively.

m:=12;
p:= func< n,x | ((x+Sqrt(x+4))^n + (x-Sqrt(x+4))^n)/2 >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( p(n,x) ), n-k) >;
[T(n,k): k in [0..n], n in [0..m]]; // G. C. Greubel, Jul 09 2023

P[n_, x_]:= P[n, x]= ((x+Sqrt[x+4])^n + (x-Sqrt[x+4])^n)/2;
T[n_, k_]:= Coefficient[Series[P[n, x], {x,0,n-k+1}], x, n-k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 08 2020; Jul 09 2023 *)

def p(n,x): return ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( p(n,x) ).list()[n-k]
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 09 2023

P(n,x) = 2*x*P(n-1,x) - (x^2 -x -4)*P(n-2,x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)] ( ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2 ).
T(n, 1) = A000217(n-1), n >= 1.
T(n, n) = A199572(n).
Sum_{k=0..n} T(n, k) = A084057(n).
Sum_{k=0..n} 2^k*T(n, k) = A125818(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A026150(n).
Sum_{k=0..n} (-2)^k*T(n, k) = A133343(n). (End)

A192421 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

2, 0, 3, 1, 8, 8, 28, 43, 111, 204, 466, 924, 2007, 4109, 8740, 18136, 38240, 79799, 167643, 350664, 735554, 1540104, 3228459, 6762553, 14172272, 29691368, 62217172, 130356451, 273144327, 572305140, 1199164498, 2512579140, 5264623167, 11030890949

Offset: 0

Clark Kimberling, Jun 30 2011

nonn

The polynomial p(n,x) is defined by ((x+d)/2)^n + ((x-d)/2)^n, where d = sqrt(x^2+4). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.

			The first five polynomials p(n,x) and their reductions are as follows:
  p(0,x) = 2 -> 2.
  p(1,x) = x -> x.
  p(2,x) = 2 + x^2 -> 3 + x.
  p(3,x) = 3*x + x^3 -> 1 + 5*x.
  p(4,x) = 2 + 4*x^2 + x^4 -> 8 + 7*x.
From these, read a(n) = (2, 0, 3, 1, 8, ...) and A192422 = (0, 1, 1, 5, 7, ...).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1).

Cf. A162514, A192232, A192422.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (2-2*x-3*x^2)/(1-x-3*x^2+x^3+x^4) )); // G. C. Greubel, Jul 11 2023

q[x_]:= x+1; d= Sqrt[x^2+4];
p[n_, x_]:= ((x+d)/2)^n + ((x-d)/2)^n (* A162514 *)
Table[Expand[p[n, x]], {n, 0, 6}]
reductionRules= {x^y_?EvenQ-> q[x]^(y/2), x^y_?OddQ-> x*q[x]^((y-1)/2)};
t = Table[FixedPoint[Expand[#1 /. reductionRules] &, p[n, x]], {n, 0, 30}]
Table[Coefficient[Part[t, n], x, 0], {n,30}] (* A192421 *)
Table[Coefficient[Part[t, n], x, 1], {n,30}] (* A192422 *)
LinearRecurrence[{1,3,-1,-1}, {2,0,3,1}, 40] (* G. C. Greubel, Jul 11 2023 *)

@CachedFunction
def a(n): # a = A192421
    if (n<4): return (2,0,3,1)[n]
    else: return a(n-1) +3*a(n-2) -a(n-3) -a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Jul 11 2023

From Colin Barker, May 12 2014: (Start)
a(n) = a(n-1) + 3*a(n-2) - a(n-3) - a(n-4).
G.f.: (2-2*x-3*x^2)/(1-x-3*x^2+x^3+x^4). (End)
a(n) = Sum_{j=0..n} T(n, j)*Fibonacci(j-1), where T(n, k) = [x^k] ((x + sqrt(x^2+4))^n + (x - sqrt(x^2+4))^n)/2^n. - G. C. Greubel, Jul 11 2023

A192426 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

2, 0, 5, 1, 18, 13, 81, 106, 413, 729, 2258, 4653, 12833, 28666, 74493, 173545, 437346, 1041421, 2583089, 6221322, 15304541, 37079289, 90826994, 220729069, 539487297, 1313161498, 3205831869, 7809748489, 19054635650, 46439068365

Offset: 0

Clark Kimberling, Jun 30 2011

The polynomial p(n,x) is defined by ((x+d)/2)^n + ((x-d)/2)^n, where d = sqrt(x^2+8). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.

			The first five polynomials p(n,x) and their reductions are as follows:
  p(0,x) = 2 -> 2
  p(1,x) = x -> x
  p(2,x) = 4 + x^2 -> 5 + x
  p(3,x) = 6*x + x^3 -> 1 + 8*x
  p(4,x) = 8 + 8*x^2 + x^4 -> 18 + 11*x.
From these, read a(n) = (2, 0, 5, 1, 18, ...) and A192427 = (0, 1, 1, 8, 11, ...).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-2,-4).

Cf. A000045, A162514, A192232, A192427.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (2-2*x-5*x^2)/(1-x-5*x^2+2*x^3+4*x^4) )); // G. C. Greubel, Jul 12 2023

q[x_]:= x+1; d= Sqrt[x^2+8];
p[n_, x_]:= ((x+d)/2)^n + ((x-d)/2)^n (* suggested by A162514 *)
Table[Expand[p[n, x]], {n, 0, 6}]
reductionRules = {x^y_?EvenQ -> q[x]^(y/2), x^y_?OddQ -> x q[x]^((y - 1)/2)};
t= Table[FixedPoint[Expand[#1/. reductionRules] &, p[n,x]], {n,0,30}]
Table[Coefficient[Part[t, n], x, 0], {n,30}] (* A192426 *)
Table[Coefficient[Part[t, n], x, 1], {n,30}] (* A192427 *)
LinearRecurrence[{1,5,-2,-4}, {2,0,5,1}, 40] (* G. C. Greubel, Jul 12 2023 *)

@CachedFunction
def a(n): # a = A192426
    if (n<4): return (2,0,5,1)[n]
    else: return a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Jul 12 2023

From Colin Barker, May 12 2014: (Start)
a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4).
G.f.: (2-2*x-5*x^2)/(1-x-5*x^2+2*x^3+4*x^4). (End)
a(n) = Sum_{k=0..n} T(n, k)*Fibonacci(k-1), where T(n, k) = [x^k] ((x + sqrt(x^2+8))^n + (x - sqrt(x^2+8))^n)/2^n. - G. C. Greubel, Jul 12 2023

Typo in name corrected by G. C. Greubel, Jul 12 2023

A195662 Triangle T(n,k) read by rows: T(0,0)= -3, T(1,0)= 2, T(1,1) = 0 and T(n,k) = T(n-1,k) -T(n-2,k-2) otherwise.

Original entry on oeis.org

-3, 2, 0, 2, 0, 3, 2, 0, 1, 0, 2, 0, -1, 0, -3, 2, 0, -3, 0, -4, 0, 2, 0, -5, 0, -3, 0, 3, 2, 0, -7, 0, 0, 0, 7, 0, 2, 0, -9, 0, 5, 0, 10, 0, -3, 2, 0, -11, 0, 12, 0, 10, 0, -10, 0, 2, 0, -13, 0, 21, 0, 5, 0, -20, 0, 3, 2, 0, -15, 0, 32, 0, -7, 0, -30, 0, 13, 0

Offset: 0

Paul Curtz, Sep 22 2011

In the notation of A195673, this defines polynomials P(n,x,p=-3,q=2), where p and q are the values of the constant and linear order for n=0 and 1.
Row sums -- the value P(n,1,-3,2) of the polynomial -- are A130848(n+5).
For general seed values in the two top rows of the triangle, the recurrence T(n,k) = T(n-1,k) - T(n-2,k-2) defines the triangle
p;
q,   0;
q,   0,      -p;
q,   0,    -p-q, 0;
q,   0,   -p-2q, 0,     p;
q,   0,   -p-3q, 0,  2p+q,  0;
and a companion triangle by adding 1 to both seed values:
p+1;
q+1, 0;
q+1, 0,    -p-1;
q+1, 0,  -p-q-2, 0;
q+1, 0, -p-2q-3, 0,    p+1;
q+1, 0, -p-3q-4, 0, 2p+q+3, 0;
The point-by-point difference between two companions is P(n,x,p+1,q+1) - P(n,x,p,q) = S(n,x) as represented (with increasing exponents) by A053119.
Examples of such triangles are A053119 (p=q=1), A192575 (p=1, q=2),
A162514 (p=2, q=1, up to a sign factor), A192011 (p=-1, q=2), A135929 (p=-2, q=1, apart from a irregular leading T(0,0)).

			The first few rows are
-3;
2, 0;
2, 0,   3;
2, 0,   1, 0;
2, 0,  -1, 0, -3;
2, 0,  -3, 0, -4, 0;
2, 0,  -5, 0, -3, 0,  3;
2, 0,  -7, 0,  0, 0,  7, 0;
2, 0,  -9, 0,  5, 0, 10, 0,  -3;
2, 0, -11, 0, 12, 0, 10, 0, -10, 0;
2, 0, -13, 0, 21, 0,  5, 0, -20, 0, 3;

Cf. A135929, A192011.

p = -3; q = 2; t[0, 0] = p; t[, 0] = q; t[, ?OddQ] = 0; t[n, k_] /; k > n = 0; t[n_ /; n >= 0, k_ /; k >= 0] := t[n, k] = t[n-1, k] - t[n-2, k-2]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2012 *)

T(n,0) = 2 (n>0).
T(n,2) = -A060747(n-3), n>2.
T(n,4) = A028347(n-5), n>6.
T(2n,2n) = -3*(-1)^n ; T(n, 2k-1) = 0 ; T(2n+1,2n) = -(3n-2)*(-1)^n.  - M. F. Hasler, Sep 28 2011

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